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Math Help - Quadratic Equation - Nature of roots

  1. #1
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    Quadratic Equation - Nature of roots

    (a)Find the value of m for which the line y=mx-3 is a tangent to the curve y=x+\frac{1}{x} and find the x-coordinate of the point at which this tangent touches the curve.

    (b) Find the value of c and of d for which (x:-5<x<3) is the solution set of  x^2+cx<d

    For (a), how do I actually solve the quadratic equation when the equation have a x^{-1}?

    No idea how to solve (b)... The whole question is just like alien to me....
    Last edited by Punch; January 17th 2010 at 06:13 AM.
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  2. #2
    Super Member Bacterius's Avatar
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    This has nothing to do with quadratic equations. This is not algebra but calculus.
    You have to take the derivative y' of y :

    y = x + \frac{1}{x} = x + x^{-1}

    Thus, following first principles, y' = 1 - x^{-2}

    Have you done derivatives ? Can you see how to do it now ?
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  3. #3
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    Quote Originally Posted by Punch View Post
    (a)Find the value of m for which the line y=mx-3 is a tangent to the curve y=x+\frac{1}{x} and find the x-coordinate of the point at which this tangent touches the curve.

    (b) Find the value of c and of d for which (x:-5<x<3) is the solution set of  x^2cx<d

    For (a), how do I actually solve the quadratic equation when the equation have a x^{-1}?

    No idea how to solve (b)... The whole question is just like alien to me....
    hi punch ,

    for (1) , you multiply the whole thing by x so you don end up with x^{-1}

    mx-3=x+\frac{1}{x}

    mx^2-3x=x^2+1

    (m-1)x^2-3x-1=0 **

    If a line is a tangent to the curve , it touches the curve at one point .

    so b^2-4ac=0

    (-3)^2-4(m-1)(-1)=0

    calculate the value of m , then put this value of m into ** , evaluate the x-coordinate then .

    for (2) , you are given the range , so

    (x+5)(x-3)<0\Rightarrow x^2+2x-15<0

    compared this with , x^2+cx-d<0 ?? (i think you have a typo here)

    we see that c=2 , d=15 assuming that i guess it correctly .
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  4. #4
    Super Member Bacterius's Avatar
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    Forgive my last post, I guess there must be a way to do it with algebra ... the hard way.
    Forgive also what I said in this post, lol, I must be tired.
    As mathaddict said, try to multiply the equations to get rid of the fractions and revert back to the good old quadratic equation. It makes it a lot easier.
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  5. #5
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    Quote Originally Posted by mathaddict View Post
    hi punch ,

    for (1) , you multiply the whole thing by x so you don end up with x^{-1}

    mx-3=x+\frac{1}{x}

    mx^2-3x=x^2+1

    (m-1)x^2-3x-1=0 **

    If a line is a tangent to the curve , it touches the curve at one point .

    so b^2-4ac=0

    (-3)^2-4(m-1)(-1)=0

    calculate the value of m , then put this value of m into ** , evaluate the x-coordinate then .

    for (2) , you are given the range , so

    (x+5)(x-3)<0\Rightarrow x^2+2x-15<0

    compared this with , x^2+cx-d<0 ?? (i think you have a typo here)

    we see that c=2 , d=15 assuming that i guess it correctly .
    yes sorry for my typo. Well, thanks and I understood part a fully. About part b, it is about turning the x values into equation and then expanding and comparing to find c and d... it just seems so much like any other quadratic equations thanks
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  6. #6
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    Quote Originally Posted by Punch View Post
    (a)Find the value of m for which the line y=mx-3 is a tangent to the curve y=x+\frac{1}{x} and find the x-coordinate of the point at which this tangent touches the curve.

    (b) Find the value of c and of d for which (x:-5<x<3) is the solution set of  x^2+cx<d

    For (a), how do I actually solve the quadratic equation when the equation have a x^{-1}?
    What quadratic equation are you talking about? There is no quadratic equation here. A line crosses a graph if the equation of the line and the equation of the graph have a common factor- that is, if mx- 3= x+ \frac{1}{x}. The line is tangent to the curve if that is a double root. To get a quadratic equation, multiply both sides of the equation by x.

    For (b), start by finding solutions to the equation x^2+ cx= d. Those two points must separate "<" from ">". Try a single point, say x= 0, to determine which is which.
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  7. #7
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    Sorry. the "quadratc equation i was talking about was an equation with x to the power of -1, the question have been solved.
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